Work-Energy problem involving pulley

In summary: It is not on a table, there is a pulley. ehildBut the normal force is not relevant for B anyway. You must take into account the acceleration of B when drawing its free body diagram. ehildOkay, so B has the tension of 3T pulling it upwards due to the 3 strings, and the force of gravity pulling it downwards. But how does that connect to the free body diagram?So the free body diagram for B has only two forces: the weight, acting downwards, and the tension in the string, acting upwards: ehildOkay, so B has the tension of 3T pulling it upwards due to the 3 strings, and the force of gravity pulling it downwards.
  • #1
thepatient
164
0

Homework Statement


The system shown is at rest when a constant 250N force is applied to block A. Neglecting masses of pulleys and friction, determine:
a) Velocity of block B after block A moved 2 m.
b) The tension in the cable
http://imageshack.us/photo/my-images/259/photoivl.jpg/

Homework Equations



Fnet = ma.

U = mgh

KE = 1/2 mv^2

Ui + KEi + W = Uf + KEf, where subindex i and f mean initial and final states.

The Attempt at a Solution



I have completed this problem, but I'm not sure if this is correct. The professor said this particular problem can be done by taking both masses and the ropes as a single system and also by taking each mass as its own, separate system. I did took each as a separate system.

First, I drew a free body diagram for block A. The net force in the x direction, assuming left is negative gives me:

F - T = ma_A
1------------------250N - T = ma_A

The relationship between the position of block A and block B was obtained and determined to be:
X_A + 3X_B = constant. Taking first time derivative:
V_A + 3V_B = 0
V_A = - 3V_B. Second time derivative:
2--------------a_A = -3a_B

Using Newton's law for the y direction on block B:
up positive: 3T - W = m*a_B
3-----------3T - mg = m*a_B

Solving 1 for acceleration a_A, then using equation 2 to leave as a_B = form:
250-T/m_a = -3a_B
4----- -250+T/(3m_a) = a_B

Using 4 to solve for tension in 3:
3T - m_b*g = -m_b*250/(3*m_a) + m_b * T/(3m_a)

Subsituting for values of mass of b, g, mass of a and solving for T:
T(3-25/90) = 25*9.81 - 25*250/(3*30)
T = 64.6N


Then, with that tension, using the work energy theorem on block A:

Wtension + Wconst force = 1/2 m_a V_a^2
250*2 - 129.2 = 1/2 *30va^2
V_a = 4.97 m /s <------ left.

Using relationship:
V_a = -3Vb
V_b = 1.66 m / s Upwards.





Does that seem right...? Thanks a ton. :]
 
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  • #2
Both bodies move along the rope, there is no reason to distinguish between x and y components of velocity. When A moves to the left, B will move vertically upward. Taking "to the left" as positive direction for A, "up" is positive for B. You correctly figured out that the acceleration of A is three times the acceleration of B. So aA=3aB.
The equations for the accelerations and forces you wrote are correct with this choice of the direction of accelerations:

25-T=mAaA
and
3T-mBg=mBaB

Proceed from here, but do not forget the parentheses. You had at least one error by ignoring them.

ehild
 
  • #3
I tried it again and more carefully this time, but used only the magnitudes of acceleration in F =ma equations. I ended up getting a tension of about 96N I think. I can try the calculation here on my phone. XD

250-T = 25a_a
3T - 30g = 30 a_b

a_A = 3a_B

Plugging 3a_B in first equation:
250 - T = 75a_B

1/75 (250 - T) = a_B =1/30(3T - 30g)
250/75 -T/75 = T/10 - g

T/75 + T/10 = 250/75 + g
T + 7.5T = 250 + 75g
Oh great now I get T = 116N :/
 
  • #4
No wait I think that's what I got for my original calculation actually. And velocity of block A comes out to be 4.63 m/s to the left using the W-E theorem and the speed of block B a third of that value.
 
  • #5
thepatient said:
I tried it again and more carefully this time, but used only the magnitudes of acceleration in F =ma equations. I ended up getting a tension of about 96N I think. I can try the calculation here on my phone. XD

250-T = 25a_a
3T - 30g = 30 a_b

The masses are interchanged in the equations above: ma = 30 kg and mb=25 kg. Your first calculation was correct, T=95.9 N.
 
  • #6
thepatient said:
No wait I think that's what I got for my original calculation actually. And velocity of block A comes out to be 4.63 m/s to the left using the W-E theorem and the speed of block B a third of that value.


I got va=4.63 m/s, vb=1.51 m/s, but it seems all right otherwise.

ehild
 
  • #7
Thanks a ton. :]
 
  • #8
You are welcome. It was a nice solution of a nice problem:smile:


ehild
 
  • #9
thepatient can you explain how you arrived at 3T-W=mbab ?
 
  • #11
  • #12
It is the weight downwards and three times the tension in the string upwards, as the block is held by three threads of the string.
What do you mean on "the normal of its weight"?

ehild
 
  • #13
Uh sorry my bad, I was referring to the normal reaction on an object kept on a table lol xD here it doesn't apply sorry
 

FAQ: Work-Energy problem involving pulley

1. What is the work-energy problem involving a pulley?

The work-energy problem involving a pulley is a physics problem that involves calculating the work done and energy transferred in a system that includes a pulley. It often involves objects with different masses connected by a rope or string that is routed through a pulley.

2. How is work calculated in a pulley system?

In a pulley system, work is calculated by multiplying the force applied to the rope by the distance the object moves. This force is equal to the weight of the object on one side of the pulley. If the object is moving up, the force is equal to the weight minus the tension in the rope. If the object is moving down, the force is equal to the weight plus the tension in the rope.

3. What is the equation for calculating the work done in a pulley system?

The equation for calculating work in a pulley system is W = Fd, where W is work, F is force, and d is distance moved. However, in a more complex pulley system, the equation may be W = (F1 + F2)(d1 + d2) where F1 and F2 are the forces acting on each side of the pulley and d1 and d2 are the distances moved by each object.

4. How is energy transferred in a pulley system?

In a pulley system, energy is transferred from one object to another through the tension in the rope. As one object moves, the other object's potential energy decreases and its kinetic energy increases. The energy transferred is equal to the work done on the object.

5. Can the work-energy problem involving a pulley be solved using conservation of energy?

Yes, the work-energy problem involving a pulley can be solved using the principle of conservation of energy. This principle states that energy cannot be created or destroyed, only transferred from one form to another. In a pulley system, the total energy at the beginning (potential energy of one object plus kinetic energy of the other) is equal to the total energy at the end (potential energy of the other object plus kinetic energy of the first). By setting these two equations equal to each other, the problem can be solved for an unknown variable.

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